Question

根据文档，我们无法从后台线程更新UI。但就我而言，我可以更新UI而不出现任何问题。检查线程3。我还打印了线程名称，结果是Thread-4。

为什么从后台线程访问主线程时不抛出异常？

fun threadsWithThreadPoolExecutor(v: View) {
    val startTime = System.currentTimeMillis()
    var endTime: Long = startTime
    val threadPoolExecutor = Executors.newFixedThreadPool(3)
    val thread1 = Thread(Runnable {
        Thread.sleep(7000)
        endTime = System.currentTimeMillis()
    })
    val thread2 = Thread(Runnable {
        Thread.sleep(10000)
        endTime = System.currentTimeMillis()
    })
    val thread3 = Thread(Runnable {
        Thread.sleep(3000)
        endTime = System.currentTimeMillis()
        tv_result.text = "This is from thread"
    })

    showProcessRunning()
    threadPoolExecutor.submit(thread1)
    threadPoolExecutor.submit(thread2)
    threadPoolExecutor.submit(thread3)

    Log.d("MyLog", thread3.name)

    Handler().postDelayed(Runnable {
        showTotalTime(startTime, endTime)
    }, 20000)
}

更新正如评论中提到的那样，我为什么在ThreadPoolExecutor＃submit（）中传递线程，因为它实现了Runnable。另外，如果我不使用ThreadPool，那么问题仍然存在。

fun threadsWithoutExecutor(v: View) {
    val startTime = System.currentTimeMillis()
    var endTime: Long = startTime
    val thread1 = Thread(Runnable {
        Thread.sleep(7000)
        endTime = System.currentTimeMillis()
    })
    val thread2 = Thread(Runnable {
        Thread.sleep(10000)
        endTime = System.currentTimeMillis()
    })
    val thread3 = Thread(Runnable {
        tv_result.text = "sdadadasda"
        Thread.sleep(3000)
        endTime = System.currentTimeMillis()
    })

    thread3.priority = Thread.MIN_PRIORITY
    showProcessRunning()
    thread1.start()
    thread2.start()
    thread3.start()

    Thread(Runnable {
        thread1.join()
        thread2.join()
        thread3.join()
        showTotalTime(startTime, endTime)
    }).start()
}

能够更新UI表单线程。为什么？

0 个答案: